Official course description: “Introduction to the concepts used in the modern treatment of solids. The student is assumed to be familiar with elementary quantum mechanics. Topics include: bonding in solids, crystal structures, lattice vibrations, free electron model of metals, band structure, thermal properties, magnetism and superconductivity (time permitting)”

This document contains:

• Plain old lecture notes. These mirror what was covered in class, possibly augmented with additional details.
• Personal notes exploring details that were not clear to me from the lectures, or from the texts associated with the lecture material.
• Assigned problems. Like anything else take these as is.
• Some worked problems attempted as course prep, for fun, or for test preparation, or post test reflection.
• Links to Mathematica workbooks associated with this course.
My thanks go to Professor Julian for teaching this course.

NOTE: This v.5 update of these notes is still really big (~18M). Some of my mathematica generated 3D images result in very large pdfs.

January 07, 2014 Two body harmonic oscillator in 3D
Figure out a general solution for two interacting harmonic oscillators, then use the result to calculate the matrix required for a 2D two atom diamond lattice with horizontal, vertical and diagonal nearest neighbour coupling.

NOTE: This v.2 update of these notes is really big (~18M), despite being only half way into the course. My mathematica generated images appear to result in very large pdfs, and I’m looking at trying pdfsizeopt to reduce the size before posting the next update (or take out the density plots from my problem set 1 solutions).

This set of notes includes the following these additions (not many of which were posted separately for this course)

Disclaimer

Question: Maximum entropy principle

Consider the “Gibbs entropy”

where is the equilibrium probability of occurrence of a microstate in the ensemble.

For a microcanonical ensemble with configurations (each having the same energy), assigning an equal probability to each microstate leads to . Show that this result follows from maximizing the Gibbs entropy with respect to the parameters subject to the constraint of

(for to be meaningful as probabilities). In order to do the minimization with this constraint, use the method of Lagrange multipliers – first, do an unconstrained minimization of the function

then fix by demanding that the constraint be satisfied.

For a canonical ensemble (no constraint on total energy, but all microstates having the same number of particles ), maximize the Gibbs entropy with respect to the parameters subject to the constraint of

(for to be meaningful as probabilities) and with a given fixed average energy

where is the energy of microstate . Use the method of Lagrange multipliers, doing an unconstrained minimization of the function

then fix by demanding that the constraint be satisfied. What is the resulting ?

For a grand canonical ensemble (no constraint on total energy, or the number of particles), maximize the Gibbs entropy with respect to the parameters subject to the constraint of

(for to be meaningful as probabilities) and with a given fixed average energy

and a given fixed average particle number

Here represent the energy and number of particles in microstate . Use the method of Lagrange multipliers, doing an unconstrained minimization of the function

then fix by demanding that the constrains be satisfied. What is the resulting ?

Answer

Writing

our unconstrained minimization requires

Solving for we have

The probabilities for each state are constant. To fix that constant we employ our constraint

or

Inserting eq. 1.15 fixes the probability, giving us the first of the expected results

Using this we our Gibbs entropy can be summed easily

or

For the “action” like quantity that we want to minimize, let’s write

for which we seek , such that

or

Our probability constraint is

or

Taking logs we have

We could continue to solve for explicitly but don’t care any more than this. Plugging back into the probability eq. 1.0.16 obtained from the unconstrained minimization we have

or

To determine we must look implicitly to the energy constraint, which is

or

The constraint () is given implicitly by this energy constraint.

Again write

The unconstrained minimization requires

or

The unit probability constraint requires

or

Our probability is then

The average energy and average number of particles are given by

The values and are fixed implicitly by requiring simultaneous solutions of these equations.

Question: Fugacity expansion ([3] Pathria, Appendix D, E)

The theory of the ideal Fermi or Bose gases often involves integrals of the form

where

denotes the gamma function.

Obtain the behavior of for keeping the two leading terms in the expansion.

For Fermions, obtain the behavior of for again keeping the two leading terms.

For Bosons, we must have (why?), obtain the leading term of for .

Answer

For we can rewrite the integrand in a form that allows for series expansion

For the th power of in this series our integral is

Putting everything back together we have for small

We’ll expand about , writing

The integral has been split into two since the behavior of the exponential in the denominator is quite different in the ranges. Observe that in the first integral we have

Since this term is of order 1, let’s consider the difference of this from , writing

or

This gives us

Now let’s make a change of variables in the first integral and in the second. This gives

As gets large in the first integral the integrand is approximately . The exponential dominates this integrand. Since we are considering large , we can approximate the upper bound of the integral by extending it to . Also expanding in series we have

For the remaining integral, Mathematica gives

where for

This gives

or

or

Evaluating the numerical portions explicitly, with

so to two terms (), we have

In order for the Boson occupation numbers to be non-singular we require less than all . If that lowest energy level is set to zero, this is equivalent to . Given this restriction, a substitution is convenient for investigation of the case. Following the text, we'll write

For , this is integrable

so that

Taylor expanding we have

Noting that , we have for the limit

or

For values of , the denominator is

To first order this gives us

Of this integral Mathematica says it can be evaluated for , and has the value

From [1] 6.1.17 we find

with which we can write

Question: Nuclear matter ([2], prob 9.2)

Consider a heavy nucleus of mass number . i.e., having total nucleons including neutrons and protons. Assume that the number of neutrons and protons is equal, and recall that each of them has spin- (so possessing two spin states). Treating these nucleons as a free ideal Fermi gas of uniform density contained in a radius , where , calculate the Fermi energy and the average energy per nucleon in MeV.

Answer

Our nucleon particle density is

With for the mass of either the proton or the neutron, and , the Fermi energy for these particles is

With , and for either the proton or the neutron, this is

This gives us

In lecture 16

we found that the total average energy for a Fermion gas of particles was

so the average energy per nucleon is approximately

Question: Neutron star ([2], prob 9.5)

Model a neutron star as an ideal Fermi gas of neutrons at moving in the gravitational field of a heavy point mass at the center. Show that the pressure obeys the equation

where is the gravitational constant, is the distance from the center, and is the density which only depends on distance from the center.

Answer

In the grand canonical scheme the pressure for a Fermion system is given by

The kinetic energy of the particle is adjusted by the gravitational potential

Differentiating eq. 1.75 with respect to the radius, we have

Noting that is the average density of the particles, presumed radial, we have

Disclaimer

This lecture had a large amount of spoken content not captured in these notes. Reference to section 4 [1] was made for additional details.

Grand canonical ensemble

Fig 1.1: Ensemble pictures

We are now going to allow particles to move to and from the system and the reservoir. The total number of states in the system is

so for , and , we have

where the chemical potential \index{chemical potential} and temperature \index{temperature} are defined respectively as

With as the set of all possible configuration pairs , we define the grand partition function

So that the probability of finding a given state with energy and particle numbers is

For a classical system we have

whereas in a quantum content we have

We want to do this because the calculation of the number of states

can quickly become intractable. We want to go to the canonical ensemble was because the partition function

yields the same results, but can be much easier to compute. We have a similar reason to go to the grand canonical ensemble, because this computation, once we allow the number of particles to vary also becomes very hard.

We are now going to define a notion of equilibrium so that it includes

All forces are equal (mechanical equilibrium)

Temperatures are equal (no net heat flow)

Chemical potentials are equal (no net particle flow)

We’ll isolate a subsystem, containing a large number of particles fig. 1.2.

Fig 1.2: A subsystem to and from which particle motion is allowed

When we think about Fermions we have to respect the “Pauli exclusion” principle \index{Pauli exclusion principle}.

Suppose we have just a one dimensional Fermion system for some potential as in fig. 1.3.

Fig 1.3: Energy level filling in a quantum system

For every momentum there are two possible occupation numbers

our partition function is

We’d find that this calculation with this constraint becomes essentially impossible.

We’ll see that relaxing this constraint will allow this calculation to become tractable.

Disclaimer

Consider a toy model of a polymer in one dimension which is made of steps (amino acids) of unit length, going left or right like a random walk. Let one end of this polymer be at the origin and the other end be at a point (viz. the rms size of the polymer) , so . We have previously calculated the number of configurations corresponding to this condition (approximate the binomial distribution by a Gaussian).

Part a

Using this, find the entropy of this polymer as . The free energy of this polymer, even in the absence of any other interactions, thus has an entropic contribution, . If we stretch this polymer, we expect to have fewer available configurations, and thus a smaller entropy and a higher free energy.

Part b

Find the change in free energy of this polymer if we stretch this polymer from its end being at to a larger distance .

Part c

Show that the change in free energy is linear in the displacement for small , and hence find the temperature dependent “entropic spring constant” of this polymer. (This entropic force is important to overcome for packing DNA into the nucleus, and in many biological processes.)

Typo correction (via email):
You need to show that the change in free energy is quadratic in the displacement , not linear in . The force is linear in . (Exactly as for a “spring”.)

Answer

Entropy.

In lecture 2 probabilities for the sums of fair coin tosses were considered. Assigning to the events for heads and tails coin tosses respectively, a random variable for the total of such events was found to have the form

For an individual coin tosses we have averages , and , so the central limit theorem provides us with a large Gaussian approximation for this distribution

This fair coin toss problem can also be thought of as describing the coordinate of the end point of a one dimensional polymer with the beginning point of the polymer is fixed at the origin. Writing for the total number of configurations that have an end point at coordinate we have

From this, the total number of configurations that have, say, length , in the large Gaussian approximation, is

The entropy associated with a one dimensional polymer of length is therefore

Writing for this constant the free energy is

Change in free energy.

At constant temperature, stretching the polymer from its end being at to a larger distance , results in a free energy change of

If is assumed small, our constant temperature change in free energy is

Temperature dependent spring constant.

I found the statement and subsequent correction of the problem statement somewhat confusing. To figure this all out, I thought it was reasonable to step back and relate free energy to the entropic force explicitly.

Consider temporarily a general thermodynamic system, for which we have by definition free energy and thermodynamic identity respectively

The differential of the free energy is

Forming the wedge product with , we arrive at the two form

This provides the relation between free energy and the “pressure” for the system

For a system with a constant cross section , , so the force associated with the system is

or

Okay, now we have a relation between the force and the rate of change of the free energy

Consider a set of independent classical harmonic oscillators, each having a frequency .

Part a

Find the canonical partition at a temperature for this system of oscillators keeping track of correction factors of Planck constant. (Note that the oscillators are distinguishable, and we do not need correction factor.)

Part b

Using this, derive the mean energy and the specific heat at temperature .

Part c

For quantum oscillators, the partition function of each oscillator is simply where are the (discrete) energy levels given by , with . Hence, find the canonical partition function for independent distinguishable quantum oscillators, and find the mean energy and specific heat at temperature .

Part d

Show that the quantum results go over into the classical results at high temperature , and comment on why this makes sense.

Part e

Also find the low temperature behavior of the specific heat in both classical and quantum cases when .

Answer

Classical partition function

For a single particle in one dimension our partition function is

with

we have

So for distinguishable classical one dimensional harmonic oscillators we have

Classical mean energy and heat capacity

From the free energy

we can compute the mean energy

or

The specific heat follows immediately

Quantum partition function, mean energy and heat capacity

For a single one dimensional quantum oscillator, our partition function is

Classical limits

matching the classical result of eq. 1.0.23. Similarly from the quantum specific heat result of eq. 1.0.31, we have

This matches our classical result from eq. 1.0.24. We expect this equivalence at high temperatures since our quantum harmonic partition function eq. 1.0.26 is approximately

This differs from the classical partition function only by this factor of . While this alters the free energy by , it doesn’t change the mean energy since . At high temperatures the mean energy are large enough that the quantum nature of the system has no significant effect.

Low temperature limits

For the classical case the heat capacity was constant (), all the way down to zero. For the quantum case the heat capacity drops to zero for low temperatures. We can see that via L’hopitals rule. With the low temperature limit is

We also see this in the plot of fig. 1.2.

Fig 1.2: Specific heat for N quantum oscillators

Question: Quantum electric dipole (2013 problem set 5, p3)

A quantum electric dipole at a fixed space point has its energy determined by two parts – a part which comes from its angular motion and a part coming from its interaction with an applied electric field . This leads to a quantum Hamiltonian

where is the moment of inertia, and we have assumed an electric field . This Hamiltonian has eigenstates described by spherical harmonics , with taking on possible integral values, . The corresponding eigenvalues are

(Recall that is the total angular momentum eigenvalue, while is the eigenvalue corresponding to .)

Part a

Schematically sketch these eigenvalues as a function of for .

Part b

Find the quantum partition function, assuming only and contribute to the sum.

Part c

Using this partition function, find the average dipole moment as a function of the electric field and temperature for small electric fields, commenting on its behavior at very high temperature and very low temperature.

Part d

Estimate the temperature above which discarding higher angular momentum states, with , is not a good approximation.

Answer

Sketch the energy eigenvalues

Let’s summarize the values of the energy eigenvalues for before attempting to plot them.

For , the azimuthal quantum number can only take the value , so we have

For we have

so we have

For we have

so we have

These are sketched as a function of in fig. 1.3.

Fig 1.3: Energy eigenvalues for l = 0,1, 2

Partition function

Our partition function, in general, is

Dropping all but terms this is

or

Average dipole moment

For the average dipole moment, averaging over both the states and the partitions, we have

For the cap of we have

or

This is plotted in fig. 1.4.

Fig 1.4: Dipole moment

For high temperatures or , expanding the hyperbolic sine and cosines to first and second order respectively and the exponential to first order we have

Our dipole moment tends to zero approximately inversely proportional to temperature. These last two respective approximations are plotted along with the all temperature range result in fig. 1.5.

Fig 1.5: High temperature approximations to dipole moments

For low temperatures , where we have

Provided the electric field is small enough (which means here that ) this will look something like fig. 1.6.

Fig 1.6: Low temperature dipole moment behavior

Approximation validation

In order to validate the approximation, let’s first put the partition function and the numerator of the dipole moment into a tidier closed form, evaluating the sums over the radial indices . First let’s sum the exponentials for the partition function, making an

With a substitution of , we have

Now we can sum the azimuthal exponentials for the dipole moment. This sum is of the form

With , and , we have

we have

With a little help from Mathematica to simplify that result we have

We can now express the average dipole moment with only sums over radial indices

So our average dipole moment is

The hyperbolic sine in the denominator from the partition function and the difference of hyperbolic sines in the numerator both grow fast. This is illustrated in fig. 1.7.

Fig 1.7: Hyperbolic sine plots for dipole moment

Let’s look at the order of these hyperbolic sines for large arguments. For the numerator we have a difference of the form

For the hyperbolic sine from the partition function we have for large

While these hyperbolic sines increase without bound as increases, we have a negative quadratic dependence on in the contribution to these sums, provided that is small enough we can neglect the linear growth of the hyperbolic sines. We wish for that factor to be large enough that it dominates for all . That is

or

Observe that the RHS of this inequality, for satisfies

So, for small electric fields, our approximation should be valid provided our temperature is constrained by

Impressed with the clarity of Baez’s entropic force discussion on differential forms [1], let’s use that methodology to find all the possible identities that we can get from the thermodynamic identity (for now assuming is fixed, ignoring the chemical potential.)

This isn’t actually that much work to do, since a bit of editor regular expression magic can do most of the work.

Our starting point is the thermodynamic identity

or

It’s quite likely that many of the identities that can be obtained will be useful, but this should at least provide a handy reference of possible conversions.